... ..... 
From 1999 through 2004 there was an active student forum
on this page for submitting and solving math problems. Although the forum
has been discontinued due to low activity, the nice selection of problems
remains for you to enjoy. All are original questions created by high school
students. The problems from 2004 appear here, with answers below. The
others can be accessed by the links provided. We also encourage students
who would like to share some of their own posers to visit The
Art of Problem Solving, where there are lively forums for posting
and solving original problems.
Previous years:...2003...2002...2001...2000...1999
...

The Problems:
......


Question 20041
submitted by Thomas Belulovich 

...... 

Question 20042
suggested by Thomas Pollom 

...... 

Question 20043
submitted by Jesse Geneson 

...... 

Question 20044
submitted by Thomas Mildorf 

...... 

Question 20045
submitted by Luke Joyner 

...... 
ANSWERS TO THE 2004 PROBLEMS
Question 20041 One way to
tackle this one is to use the fact that F(1) + F(2) + ... + F(n)
= F(n+2)  1 repeatedly. You should find that the overall sum
boils down to F(n+4)n3
Question 20042 It can be done with
just nine pentagons. Try starting with one pentagon at each corner, and
work your way towards the center.
Question 20043 The answer is 4, which
can be proved by resorting to inequalities.
Question 20044 The answer is 103/94. A brief solution posted
at the Art of Problem Solving web site appears below.
Use standard notation (O is the circumcenter, N the
ninepoint center, R the circumradius, r the inradius). We know 2 OI^2
= 2R^2  4Rr (eqn 1) and that cos A + cos B + cos C = 1 + r/R (eqn 2).
We also know that the incircle and the ninepoint circle are tangent,
so IN = R/2  r, or 4*IN^2 = R^2  4Rr + 4r^2 (eqn 3). Now, we know
NOHG are collinear on the Euler line, and that the homothecy about G
with factor 1/2 takes H to O and O to N, so GO = 1/2 * GH = 2, GN =
1/2 * GO = 1. Now, use Stewart's theorem to find IN, IO. We get 2^2
* 1 + 3^2 * 3 = 3 * 1 * 4 + 4 IN^2 => 4 IN^2 = 19 (eqn 4), and 3^2
* 6 + 2^2 * 2 = 4 * 6 * 2 + 4 IO^2 => 47 = 2 IO^2 (eqn 5). Now,
consider 47(eqn 4)  19(eqn 5). We get 47 (4 IN^2)  19 (2 IO^2) = 0.
We plug in (eqn 1) and (eqn 3) to find 47 (R^2  4Rr + 4r^2)  19 (2R^2
 4Rr) = 0 => 9R^2  112Rr + 188r^2 = 0 => 188(r/R)^2  112(r/R)
+ 9 = 0 => (2 * (r/R)  1) * (94 * (r/R)  9) = 0. The triangle's
not equilateral (otherwise all these fun distances would be 0), so we
ignore the first factor => r/R = 9/94, so from (eqn 2), we get an
answer of 103 / 94.
Question 20045 A triangle with sides
1, 2.5, and 2.5 does the job. In general the equation relating a
and b will involve a second degree polynomial in these two variables.
......
